/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 45 In Exercises \(43-46,\) compute ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Exercises \(43-46,\) compute \(F^{\prime}\) and \(F^{\prime \prime} .\) Determine the intervals on which \(F\) is increasing, decreasing, concave up, and concave down. $$ F(x)=\int_{1}^{x} t \ln (t) d t \quad 0

Short Answer

Expert verified
\( F(x) \) increases on \( x > 1 \), decreases on \( 0 < x < 1 \), is concave up on \( x > \frac{1}{e} \), and concave down on \( 0 < x < \frac{1}{e} \).

Step by step solution

01

Understanding the Problem

We need to compute the first derivative \( F'(x) \) and the second derivative \( F''(x) \) of the given integral function \( F(x)=\int_{1}^{x} t \ln(t) \, dt \). Then, determine the intervals where \( F \) is increasing or decreasing, as well as where \( F \) is concave up or concave down.
02

Compute the First Derivative Using the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, if \( F(x) = \int_{a}^{x} f(t) \ dt \), then \( F'(x) = f(x) \). Here, \( f(t) = t \ln(t) \), so \( F'(x) = x \ln(x) \).
03

Compute the Second Derivative

To find the second derivative \( F''(x) \), we differentiate \( F'(x) = x \ln(x) \) with respect to \( x \). Apply the product rule: \( d(uv)/dx = u'v + uv' \) where \( u = x \) and \( v = \ln(x) \). Thus, \( F''(x) = (1) \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1 \).
04

Determine Intervals of Increase and Decrease

\( F(x) \) increases where \( F'(x) > 0 \) and decreases where \( F'(x) < 0 \). Since \( F'(x) = x \ln(x) \), analyze \( x \ln(x) \). \( x \ln(x) > 0 \) for \( x > 1 \) and \( x \ln(x) < 0 \) for \( 0 < x < 1 \). Thus, \( F(x) \) is increasing for \( x > 1 \) and decreasing for \( 0 < x < 1 \).
05

Determine Intervals of Concavity

\( F(x) \) is concave up where \( F''(x) > 0 \) and concave down where \( F''(x) < 0 \). From Step 3, \( F''(x) = \ln(x) + 1 \). \( F''(x) > 0 \) when \( \ln(x) > -1 \) which is when \( x > \frac{1}{e} \), and \( F''(x) < 0 \) when \( x < \frac{1}{e} \). Thus, \( F(x) \) is concave up for \( x > \frac{1}{e} \) and concave down for \( 0 < x < \frac{1}{e} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative
The concept of derivatives is central to calculus. It is essentially the rate at which a function is changing at any given point. In simpler terms, it tells us how steep or flat the function's graph is at a specific point, like the speedometer of a car showing speed at an instant.

In the context of the function \( F(x) = \int_{1}^{x} t \ln(t) \, dt \), we find the first derivative \( F'(x) \) using the Fundamental Theorem of Calculus. According to the theorem, if we have an integral of the form \( F(x) = \int_{a}^{x} f(t) \, dt \), then the derivative \( F'(x) \) is simply \( f(x) \). Thus, here, \( F'(x) = x \ln(x) \).

Understanding \( F'(x) \) tells us where the function \( F(x) \) is increasing or decreasing. \( F(x) \) is increasing where \( F'(x) > 0 \) and decreasing where \( F'(x) < 0 \). For \( x \ln(x) \), it is positive when \( x > 1 \), meaning \( F(x) \) is increasing, and negative when \( 0 < x < 1 \), so \( F(x) \) is decreasing.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a bridge between differential and integral calculus. It connects the concept of an integral, which measures area, with that of a derivative, which measures slope.

This theorem states that if \( F(x) = \int_{a}^{x} f(t) \, dt \), then \( F'(x) = f(x) \). The function \( F(x) \) is called an antiderivative or an indefinite integral of \( f(x) \).

For the given problem, \( F(x) = \int_{1}^{x} t \ln(t) \, dt \), using the theorem, we immediately find that \( F'(x) = t \ln(t) \), evaluated at \( x \), or simply \( F'(x) = x \ln(x) \).

This direct replacement allows us to find a derivative of complex expressions that might otherwise require elaborate calculations. Understanding the Fundamental Theorem helps simplify solving integrals, and it's truly revolutionary in calculus for simplifying evaluations of definite integrals.
Concavity
Concavity helps us understand the shape and curvature of a function's graph. It tells us whether the graph is curving upwards or downwards:
  • A function is concave up when it curves like a cup or a smile. Mathematically, this occurs when the second derivative \( F''(x) \) is greater than 0.
  • A function is concave down when it curves like a frown. In math terms, this happens when the second derivative \( F''(x) \) is less than 0.
The second derivative test is crucial in determining concavity. In the problem, we calculated the second derivative as \( F''(x) = \ln(x) + 1 \). Using this, we can predict how \( F(x) \) behaves:
  • \( F''(x) > 0 \): \( F(x) \) is concave up for \( x > \frac{1}{e} \), because a positive second derivative indicates a curve upwards.
  • \( F''(x) < 0 \): \( F(x) \) is concave down for \( 0 < x < \frac{1}{e} \), suggesting a curve downwards.
Understanding these intervals of concavity is vital for analyzing and graphing functions, helping predict the peaks, troughs, and overall shape of the graph.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The partition \\{0,15,25,50,75,90,100\\} of [0,100] given in Exercise 46 precludes an application of Simpson's Rule because it is not uniform. However, because the number of subintervals is even, the idea behind Simpson's Rule can be used. Find the degree 2 polynomial \(p_{1}\) that passes through the three points \((0,0),(15,3),\) and \((25,19) .\) The interpolation command of a computer algebra system will make short work of finding this polynomial. Calculate \(\int_{0}^{25} p_{1}(x) d x .\) Next, find the degree 2 polynomial \(p_{2}\) that passes through the three points \((25,19),(50,25),\) and \((75,42) .\) Calculate \(\int_{25}^{75} p_{2}(x) d x .\) Repeat this procedure with the last three points, \((75,42),(90,75),\) and (100,100) . What approximation of \(\int_{0}^{100} L(x) d x\) results?

An integral \(\int_{a}^{b} f(x) d x\) and a positive integer \(N\) are given. Compute the exact value of the integral, the Simpson's Rule approximation of order \(N,\) and the absolute error \(\varepsilon\). Then find a value \(c\) in the interval \((a, b)\) such that \(\varepsilon=(b-a)^{5}\left|f^{(4)}(c)\right| /\left(180 \cdot N^{4}\right) .\) (This form of the error, which resembles the Mean Value Theorem, implies inequality \((5.8 .4) .)\) $$ \int_{1}^{e} 1 / x d x \quad N=4 $$

Find the area of the region that is bounded by the graphs of \(y=f(x)\) and \(y=g(x)\) for \(x\) between the abscissas of the two points of intersection. $$ f(x)=x^{2}+x-1 \quad g(x)=2 x+1 $$

Income data for three countries are given in the following tables. In each table, \(x\) represents a percentage, and \(L(x)\) is the corresponding value of the Lorenz function, as described in Example \(3 .\) In each of Exercises \(23-27,\) use the specified approximation method to estimate the coefficient of inequality for the indicated country. (The values \(L(0)=0\) and \(L(100)=100\) are not included in the tables, but they should be used.) $$ \begin{array}{|c|r|c|c|c|}\hline x & 20 & 40 & 60 & 80 \\\\\hline L(x) & 5 & 20 & 30 & 55\\\\\hline\end{array}$$ Income Data, Country A $$\begin{array}{|c|c|c|c|}\hline x & 25 & 50 & 75 \\\\\hline L(x) & 15 & 25 & 40 \\\\\hline\end{array}$$ Income Data, Country B $$\begin{array}{|c|r|r|r|r|r|r|r|r|r|}\hline \boldsymbol{x} & 10 & 20 & 30 & 40 & 50 & 60 & 70 & 80 & 90 \\\\\hline \boldsymbol{L}(\boldsymbol{x}) & 4 & 8 & 14 & 22 & 32 & 42 & 56 & 70 & 82 \\\\\hline\end{array}$$ Income Data, Country \(\mathbf{C}\) Country A, Trapezoidal Rule

Find the area of the region that is bounded by the graphs of \(y=f(x)\) and \(y=g(x)\) for \(x\) between the abscissas of the two points of intersection. $$ f(x)=x^{2}+x+1 \quad g(x)=2 x^{2}+3 x-7 $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.